The Stacks project

Proof. If $\dim (A) = 0$, then $U$ is empty hence affine (equal to the spectrum of the $0$ ring). If $\dim (A) = 1$, then we can choose an element $f \in \mathfrak m$ not contained in any of the finite number of minimal primes of $A$ (Algebra, Lemmas 10.31.6 and 10.15.2). Then $U = \mathop{\mathrm{Spec}}(A_ f)$ is affine.

The converse is more interesting. We will give a somewhat nonstandard proof and discuss the standard argument in a remark below. Assume $U = \mathop{\mathrm{Spec}}(B)$ is affine. Since affineness and dimension are not affecting by going to the reduction we may replace $A$ by the quotient by its ideal of nilpotent elements and assume $A$ is reduced. Set $Q = B/A$ viewed as an $A$-module. The support of $Q$ is $\{ \mathfrak m\} $ as $A_\mathfrak p = B_\mathfrak p$ for all nonmaximal primes $\mathfrak p$ of $A$. We may assume $\dim (A) \geq 1$, hence as above we can pick $f \in \mathfrak m$ not contained in any of the minimal ideals of $A$. Since $A$ is reduced this implies that $f$ is a nonzerodivisor. In particular $\dim (A/fA) = \dim (A) - 1$, see Algebra, Lemma 10.60.13. Applying the snake lemma to multiplication by $f$ on the short exact sequence $0 \to A \to B \to Q \to 0$ we obtain

where $Q[f] = \mathop{\mathrm{Ker}}(f : Q \to Q)$. This implies that $Q[f]$ is a finite $A$-module. Since the support of $Q[f]$ is $\{ \mathfrak m\} $ we see $l = \text{length}_ A(Q[f]) < \infty $ (Algebra, Lemma 10.62.3). Set $l_ n = \text{length}_ A(Q[f^ n])$. The exact sequence

shows inductively that $l_ n < \infty $ and that $l_ n \leq l_{n + 1}$. Considering the exact sequence

and we see that the image of $Q[f^ n]$ in $Q/fQ$ has length $l_ n - l_{n + 1} + l \leq l$. Since $Q = \bigcup Q[f^ n]$ we find that the length of $Q/fQ$ is at most $l$, i.e., bounded. Thus $Q/fQ$ is a finite $A$-module. Hence $A/fA \to B/fB$ is a finite ring map, in particular induces a closed map on spectra (Algebra, Lemmas 10.36.22 and 10.41.6). On the other hand $\mathop{\mathrm{Spec}}(B/fB)$ is the punctured spectrum of $\mathop{\mathrm{Spec}}(A/fA)$. This is a contradiction unless $\mathop{\mathrm{Spec}}(B/fB) = \emptyset $ which means that $\dim (A/fA) = 0$ as desired. $\square$